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PROMISCUOUS EXERCISES.

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And when x = 3 Ac, then z 886 of a foot, or 103
inches, = AE, the height of the water in the ditches when
the tide is at co or 3 feet high without, or in the first hour
and half of time.

Again, to find the time, after the above, when

EF arrives at cp, or when the water in the ditches G
arrives as high as the top of the sluice.

The notation remaining as before,

then 2bz g(x-2) per sec. runs through AF,

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B

and gb(3-2)g(x-2) per sec. thro' ED nearly; A
therefore 3bgX(12+z)√(x − z) is the whole per second
through AD nearly.

conseq.

2b √☎ × (12 + z) √ (x − z) = v is the velocity per

5A

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Assume z=Ax2+вx2+cx2+DÃ2 &c. So shall

A A2+4B 3 ▲3+4AB+8c
3

23

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√ (x − x) = x

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12+z=12+ax2+вx2+cx2 &c.;

x2 &c.;

16

(12+z). √/(x−z). x=12x3 x − 6ax3x— (§ A2+6в)x3⁄4x &c. ;

1

3

6A3

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Then, equating the like terms, &c. we have

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But by the first process, when x=3, z=•886; which substitated for them, we have z = •886, and the series 1.63: therefore the correct fluents are

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And when z=3= AC, it gives x = 6·369 for the height of the tide without, when the ditches are filled to the top of the sluice, or 3 feet high; which answers to 3h 11′ 4′′. .

Lastly,

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.

Lastly, to find the time of rising the remaining 3 feet above

the top of the sluice; let

*=ce the height of the tide above co,

z=CE ditto in the ditches above CD ;

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and the other dimensions as before.

G

GI

C

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A

Then/g: /EG: : 2g : 2 √✓ g (x− z) = the
velocity with which the water runs through the
whole sluice an; conseq. ad X2√g (x−z) —
18g (x-z) is the quantity per second running through the
sluice, and 18✓g
√ (x−2) the velocity of z, or the rise

A

of the water in the ditches, per second; hence v:ż::

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2 3, x = 7; the heights above the top of the sluice; answering to 6 and 10 feet above the bottom of the ditches. That is, for the water to rise to the height of 6 feet within the ditches, it is necessary for the tide to rise to 10 feet without, which just answers to 5 hours; and so long it would take to fill the ditches 6 feet deep with water, their horizontal area being 200000 square feet.

+

Further, when x = 6, then z = 2.117 the height above

*Note. The fluxional equation mi — ż√(x—2) may be integrated without series,-EDITOR,

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the top of the sluice; to which add 3, the height of the
sluice, and the sum 5.117, is the depth of water in the ditches
in 4 hours and a half, or when the tide has risen to the height
of 9 feet without the ditches..

Note. In the foregoing problems, concerning the afflux of
water, it is taken for granted that the velocity is the same as
that which is due to the whole height of the surface of the
supplying water: a supposition which agrees with the prin-
ciples of the greater number of authors: though some make
the velocity to be that which is due to the half height only:
and others make it still less.

Also in some places, where the difference between two pa-
rabolic segments was to be taken, in estimating the mean ve-
locity of the water through a variable orifice, I have used a
near mean value of the expression; which makes the opera-
tion of finding the fluents much more easy, and is at the same
time sufficiently exact for the purpose in hand.

We may further add a remark here concerning the method
of finding the fluents of the three fluxional forms that occur
in the solution of this problem, viz. the three forms mz
(2x+2) √ (x — z) x, and mż (12 + 2) √(x-2), and
mz (x-2)x the fluents of which are found by assuming
the fluent me in an infinite series ascending in terms of x with
indeterminate coefficients A, B, C, &c. which coefficients are
afterwards determined in the usual way, by equating the cor-
responding terms of two similar and equal series, the one se-
ries denoting one side of the fluxional equation, and the other
series the other side. By similar series, is meant when they
have equal or like exponents; though it is not necessary that
the exponents of all the terms should be like or pairs, but on-
ly some of them, as those that are not in pairs will be can-
celled or expelled by making their coefficients = 0 or nothing.
Now the general way to make the two series similar, is to
assume the fluent z equal to a series in terms of x,
either as-
cending or descending, as here

z = x2 + x2+s+x+2s &c. for ascending,
or z x2+x2¬3+xˆ−3 &c. for a descending
series, having the exponents r, r ±s, r ± 2s, &c. in arith-
metical progression, the first term r, and common difference
s; without the general coefficients A, B, C, &c. till the values
of the exponents be determined. In terms of this assumed
series for z, find the values of the two sides of the given
fluxional equation, by substituting in it the said series instead
of z; then put the exponent of the first term of the one side
equal that of the other, which will give the value of the first
exponent r; in like manner put the exponents of the two 2d

terms

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terms equal, which will give the value of the common differ-
ences; and hence the whole series of exponents r, r±s, r
2s &c. becomes known.

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Thus, for the last of the three fluxional equations above
mentioned, viz. mz=√(x-z)x, or only z (x − x) x;
having assumed as above z x+x+ &c. and taking the
fluxion, then z = x2−1 x + x2+-1 x + &c. omitting the co-
efficients; and the other side of the equation (− z) x =
√(xx" - x2+s &c.) xx-xr-x &c. Now the exponents
of the first terms made equal, give r—1=1, theref. r=
r=1+1
-3; and those of the 2d terms made equal, give r+s—1=r
, theref. s-1= 1, and s=1-1-1: conseq. the whole
assumed series of exponents r, rts, r+2s, &c. become,,,
&c. as assumed above.

s—1X

2

x

Again, for the 2d equation mzor z = (12 + z) √ (x − z) ∞
= (a + z) √(x-2); assuming z=x+x+s &c. as before,
then z = x2-1x + xrts-1x &c. and /(x-z) x = x x x x
13
&c. both as above; this mult. by a +z or a + x2 + x2+s &c.
gives axxax-x &c: then equating the first exponents
gives r-1 or r, and r+s-1=r-, or s=1-1=1;
hence the series of exponents is 3, 4, 4, &c. the same as the
former, and as assumed above.

Lastly, assuming the same form of series for z and z as in
the above two cases, for the 1st fluxional equation also, viz.
mz=(2x+z) √✓ (x −z)x then (x-z) x = x2x - x2-x &c.

: /

X xr

1

which mult. by 2x+z, gives 2x3x--xr+2x &c. : here equat-
ing the first exponents gives r-1=orr, and equating
the 2d exponents gives r+s 1: = r + 1, or s = 2; hence
the series of exponents in this case is 5, 9, 7,
8 11, &c. as used
for this case above. Then, in every case, the general co-
efficients A, B, c, &c. are joined to the assumed terms x, xs,
&c. and the whole process conducted as in the three series
just referred to.

Such then is the regular and legitimate way of proceeding,
to obtain the form of the series with respect to the exponents
of the terms. But in many cases we may perceive at sight,
without that formal process, what the law of the exponents
will be, as I indeed did in the solutions in the series above
referred to; and any person with a little practice may easily
do the same.

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VOL. II

71

PROBLEM

:

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PROBLEM 53.

To determine the Fall of the Water in the Arches of a Bridge.

The effects of obstacles placed in a current of water, such as the piers of a bridge, are, a sudden steep descent, and an increase of velocity in the stream of water, just under the arches, more or less in proportion to the quantity of the obstruction and velocity of the current; being very small and hardly perceptible where the arches are large and the piers few or small, but in a high and extraordinary degree at London-bridge, and some others, where the piers and the sterlings are so very large, in proportion to the arches. This is the case, not only in such streams as run always the same way, but in tide rivers also, both upward and downward, but much less in the former than in the latter. During the time of flood, when the tide is flowing upward, the rise of the water is against the under side of the piers; but the difference between the two sides gradually diminishes as the tide flows less rapidly towards the conclusion of the flood. When this has attained its full height, and there is no longer any current, but a sțillness prevails in the water for a short time, the surface assumes an equal level, both above and below bridge. But as soon as the tide begins to ebb or return again, the resistance of the piers against the stream, and the contraction of the waterway, cause a rise of the surface above and under the arches, with a full and a more rapid descent in the contracted stream just below. The quantity of this rise, and of the consequent velocity below, keep both gradually increasing, as the tide continues ebbing, till at quite low water, when the stream or natural current being the quickest, the fall under the arches is the greatest. And it is the quantity of this fall which it is the object of this problem to determine.

1

Now, the motion of free running water is the consequence of and produced by the force of gravity, as well as that of any other falling body. Hence the height due to the velocity, that is, the height to be freely fallen by any body to acquire the observed velocity of the natural stream, in the river a little way above bridge, becomes known. From the same velocity also will be found that of the increased current in the narrowed way of the arches, by taking it in the reciprocal proportion of the breadth of the river above, to the contracted way in the arches; viz. by saying, as the latter is to the former, so is the first velocity, or slower motion, to the quicker. Next, from this last velocity, will be found the height due to

it

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